Here it is:
$\displaystyle (a + b)x + cy = bc $
$\displaystyle (b + c)y + ax = -ab$
Find x and y.
I've tried this for ages but the answers are $\displaystyle x= c$ and $\displaystyle y=-a$
Please help
Here it is:
$\displaystyle (a + b)x + cy = bc $
$\displaystyle (b + c)y + ax = -ab$
Find x and y.
I've tried this for ages but the answers are $\displaystyle x= c$ and $\displaystyle y=-a$
Please help
In matrix form this would be
$\displaystyle \displaystyle \left[\begin{matrix}a+b&c\\b+c&a\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}bc\\8722ab\end{matrix}\right]$
$\displaystyle \displaystyle \left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}a+b&c\\b+c&a\end{matrix}\right]^{-1}\left[\begin{matrix}bc\\8722ab\end{matrix}\right]$.
Can you go from here?
By the substitution method:
$\displaystyle [1] \,\,\, (a + b)x + cy = bc $
$\displaystyle [2] \,\,\, (b + c)y + ax = -ab$
$\displaystyle [2] \,\,\, y = \dfrac{-ab - ax}{b + c}$
Substitute into [1]:
$\displaystyle [1] \,\,\, (a + b)x + \dfrac{c(-ab - ax)}{b + c} = bc $
And the simplify to solve for x. Place the value of x back into one of the equations to solve for y.
Hello, Prove it!
Be careful . . . The original equations were "scrambled" . . .
$\displaystyle \begin{array}{ccc}(a + b)x + cy &=& bc \\
ax + (b + c)y &=& \text{-}ab\end{array}\qquad \text{(unscrambled form)}$
We want: .$\displaystyle \begin{bmatrix}a+b & c \\ a & b+c\end{bmatrix}\,\begin{bmatrix} x \\ y\end{bmatrix} \;=\;\begin{bmatrix}bc \\ \text{-}ab\end{bmatrix}$
Huh? Try a = 2, b = 4, c = -6, x = 1, y = 5
That's a solution for both equations.....but x<>c and y<>-a
There are many other similar cases.
Did I miss something?
Also, equations can be re-arranged this way:
ax + bx + cy = bc [1]
-ax - by - cy = ab [2]
===============
bx - by = ab + bc
x - y = a + c